Is function $f:\mathbb C-\{0\}\rightarrow\mathbb C$ prescribed by $z\rightarrow \large \frac{1}{z}$ by definition discontinuous at $0$? Is function $f:\mathbb C-\{0\}\rightarrow\mathbb C$ prescribed by  $z\rightarrow \large{\frac{1}{z}}$ by definition discontinuous at $0$?
Personally I would say: "no". In my view a function can only be (dis)continuous at $z$ if $z$ belongs to its domain.
But I have heard other sounds, that made me curious. 
This question was inspired by comments/answers on this question.
 A: For a function $f$ to be continuous at a point $a$, you must have $a\in\text{dom}(f)$.  The function you cite is continuous on the punctured plane.
A: Typically continuity or lack thereof is restricted to points on the domain where the function is defined, but it can also be reasonably extended to your problem.
The question is not "is $f$ continuous at zero?" but "can $f$ be continuously extended to zero?", and this makes perfect sense.
Since we are talking about extension to a single point, the question is whether $f$ has a limit at that point.
The answer depends on the target space.
If you want $f$ to map to $\mathbb C$, there is no limit.
If you want it to map to the Riemann sphere $\mathbb C\cup\{\infty\}$, then there is.
In fact, the mapping $z\mapsto1/z$ is a bijection of the Riemann sphere to itself.
A: I suspect that there is no universal agreement among different sources. But for example Rudin's Principles (p. 94) says "If $x$ is a point in the domain of the function $f$ at which $f$ is not continuous, we say that $f$ is discontinuous at $x$, or that $f$ has a discontinuity at $x$". He doesn't mention anything about points not in the domain of $f$, but this omission sort of implies that for such points neither of the terms continuous or discontinuous should be applied.
I think this practice makes a lot of sense, since your example function is continuous (being continuous at all points in its domain), and allowing continuous functions to have discontinuities would be strange, wouldn't it? (Singularity is a better word in such a case.)
